How to calculate the integral $\int e^{\cos x}\cos (x+\sin x) dx$ With the help of Mathematica we find 
$$\int e^{\cos x}\cos (x+\sin x)\ dx = e^{\cos x}\sin (\sin x)$$
But I tried normal method like integrating by parts, without success.
 A: You can do this with real methods using the sum-angle formula to write
$$e^{\cos(x)}\cos(x+\sin(x))\\=e^{\cos(x)}\cos(x)\cos(\sin(x))-e^{\cos(x)}\sin(x)\sin(\sin(x))$$
This is now recognizably in the form $u'v+v'u=(uv)'$, and you may use the product rule of differentiation to obtain your answer.
A: Just to spell out achille hui's use of complex numbers, , your integral is $$\Re\int\exp(\cos x + i(x+\sin x))dx=\Re\int(\cos x+i\sin x)\exp(\cos x + i\sin x)dx\\=\Re(-i\exp(\cos x+i\sin x)+C)=\Im\exp (\cos x+i\sin x)+\Re C\\=\exp\cos x\cdot\sin\sin x+\Re C.$$
A: Note that $$\cos(x+\sin x)=\cos x\cos(\sin x) -\sin x\sin(\sin x).$$ Hence it holds that
$$\begin{eqnarray}
\int e^{\cos x}(\cos x\cos(\sin x) -\sin x\sin(\sin x))dx &=& \int e^{\cos x}\cos x\cos(\sin x)dx \\&&-\int e^{\cos x}\sin x\sin(\sin x)dx \\&=& \operatorname{I}-\operatorname{II}.
\end{eqnarray}$$ The first term is equal to
$$\begin{eqnarray}
\operatorname{I}&=&\int e^{\cos x}\cos x\cos(\sin x)dx=\int e^{\cos x}(\sin(\sin x))'dx \\&=&e^{\cos x}\sin(\sin x)+\int e^{\cos x}\sin x \sin(\sin x))dx =e^{\cos x}\sin(\sin x)+\operatorname{II},
\end{eqnarray}$$ by integration by parts. It follows that
$$
\operatorname{I}-\operatorname{II} = e^{\cos x}\sin(\sin x),
$$ as desired.
